# What is the formula for the first Riemann zeta zero?

I found this approximation of which an earlier version I posted in the chat room:

$$7 \pi -\text{Log}\left[\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi \right] = 14.13472514154629716253329494571302508888...$$

The first non trivial zeta zero: $$14.13472514173469379045725198356247027078$$

Can you improve on the formula above?

Edit 2.9.2012

Based on the comments below I would like to explain how I reasoned:

Any Taylor series evaluated at $x=1$ is convergent for variants of it when multiplied element wise with rows in this matrix:

$$\begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$

Many Taylor series have the second row as part of its coefficients. That is: $$(1,-1,1,-1,1,-1,1,-1,1,-1,...)$$

Such Taylor series are for example $\log 2$, $\sqrt 2$, $\cos 1$, $\sin 1$. The reason for the convergence of such series and divisibility defined variants of thereof, seems to be that in the matrix above, a period sums to zero.

The simplest Dirichlet series that sums to zero and is not a an element wise multiplication of two other Dirichlet series, is the first row:

$$\frac{0}{1}+\frac{0}{2}+\frac{0}{3}+\frac{0}{4}+\frac{0}{5}+... = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

This suggests that one should try to find an expression for a such sequence.

The definition of a number raised to a complex number is:

$$n^{(a+ib)} = n^{a}(\cos (b \log (n))+i\sin (b \log (n)))$$

and the Riemann zeta function is:

$$\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+...$$

where $s$ is a complex number.

Here I then made a mistake. I started studying the equation: $$\cos (\log (n)) = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ in order to get something similar to the Dirichlet series with numerators equal to the all zeros sequence in expression $(1)$ above. But if I understand correctly this would be the same as seeking the undefined sequence:

$$\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+\frac{1}{0}+$$

After that I just guessed that by combining values from the solutions to equation $(2)$ one could possibly find an expression for the zeta zeros.

Edit 23.12.2012: For what it is worth. Here is how the actual calculation went:

The first Riemann zeta zero is:

$$\Im(\rho _1)$$ $$=14.1347251417346937904572519836$$

A number close to the first Riemann zeta zero is:

$$\frac{9 \pi }{2}$$ $$=14.1371669411540695730818952248$$

That number can be split up into:

$$\frac{9 \pi }{2} = 7 \pi -\log \left(e^{\frac{5 \pi }{2}}\right)$$

To see what is missing within the logarithm I added an $x$ and solved the equation:

$$\text{Solve}\left[N\left[7 \pi -\log \left(x+e^{\frac{5 \pi }{2}}\right),30\right]=N[\Im(\rho _1),30],x\right]$$

This gives the solution:

$$\{\{x\to 6.297688980465813720589098\}\}$$

which is close to:

$$2\pi = 6.28318530717958647692528676656...$$

Substituting $x$ with $2\pi$:

$$7 \pi -\log \left(e^{\frac{5 \pi }{2}}+2 \pi \right)$$

which is closer:

$$=14.1347307583914370155699744066$$

Some small number seems to be missing, the second harmonic number could be it:

$$7 \pi -\log \left(e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

which again is closer:

$$=14.1347272795405950845865949010$$

Multiplying the added number with $\frac{3}{2}$

$$7 \pi -\log \left(\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

closer still:

$$=14.1347255401197125097619679160$$

continuing the trick with similar numbers:

$$7 \pi -\log \left(\frac{5}{2} e^{-\frac{1}{2} (5 \pi )}+\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

works:

$$=14.1347251642841507747886817861$$

and once more:

$$7 \pi -\log \left(\frac{7}{2} e^{-\frac{1}{2} (7 \pi )}+\frac{5}{2} e^{-\frac{1}{2} (5 \pi )}+\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

it works:

$$=14.1347251415462971625332949457$$

but then I can't get further.

Edit: 5.11.2013:

$$\frac{\sqrt{\frac{\Im(\rho _1)}{\pi }+\frac{1}{2}}}{\sqrt{5}}=0.999922272089659461895288929782$$

$$\frac{\Im(\rho _1)}{\pi }+\frac{1}{2}=4.99922275110473484848654142318$$

$\rho _1$ = first riemann zeta zero = 14.134725141734693790457...

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You might want to add some explanations about this exact-approximate formula (the one with $\pi$ and exponentials). Is it composed of the first terms of an explicit series? –  Did Sep 2 '12 at 17:01
Interesting, but $2171956\over153661$ is closer. I mean, unless the pattern lurking in your exression leads to a convergent sequence (which seems is not the case), this looks quite fruitless to me. –  Hagen von Eitzen Sep 2 '12 at 17:02
Agreed, unless you can come up with a "reason," this could easily just be a coincidence. –  Thomas Andrews Sep 2 '12 at 17:06
I will try to explain by editing the question above, with a Edit section below the original question. –  Mats Granvik Sep 2 '12 at 17:11
I fail to see what the Edit section explains. –  Did Sep 2 '12 at 18:15

But im also a bit skeptical and i will try to explain why.

In short , it may just be curve fitting.

You see you got about a few handful integers in your approximation and a few handful correct decimals.

Thus your amount of correct decimals is not much more than the length of your formula.

I hope you understand that.

On the other hand , you got the numbers e and pi involved and the integers follow a pattern.

That makes it seems nonrandom again contrary to the argument above.

Hence its an interesting question.

I wonder how you arrived at that " result ".

One can express the zero's of meromorphic functions with integrals , especially contour integrals.

Maybe if we convert such an integral in a sum we might make progress with a series for the first zero.

Although this might look different from your approximation. On the other hand if we can transform that series into another one we might get closer to your approximation.

Regards

mick

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Can you improve on the formula above? -- Yes: $$\text{Log}\left[\frac{2}{3} e^{-5 \pi /2}+\frac{e^{7 \pi }}{\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi }\right] = 14.1347251417343...$$